ZOJ 2243 Binary Search Heap Construction笛卡尔树(二叉搜索+堆)

阐述了如何构造一个带优先级的堆,并通过中序遍历实现特定数据集的组织与排序。

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Binary Search Heap Construction

Time Limit: 5 Seconds      Memory Limit: 32768 KB

Read the statement of problem G for the definitions concerning trees. In the following we define the basic terminology of heaps. A heap is a tree whose internal nodes have each assigned a priority (a number) such that the priority of each internal node is less than the priority of its parent. As a consequence, the root has the greatest priority in the tree, which is one of the reasons why heaps can be used for the implementation of priority queues and for sorting.

A binary tree in which each internal node has both a label and a priority, and which is both a binary search tree with respect to the labels and a heap with respect to the priorities, is called a treap. Your task is, given a set of label-priority-pairs, with unique labels and unique priorities, to construct a treap containing this data.

Input Specification

The input contains several test cases. Every test case starts with an integer n. You may assume that 1<=n<=50000. Then follow n pairs of strings and numbers l1/p1,...,ln/pndenoting the label and priority of each node. The strings are non-empty and composed of lower-case letters, and the numbers are non-negative integers. The last test case is followed by a zero.

Output Specification

For each test case output on a single line a treap that contains the specified nodes. A treap is printed as (<left sub-treap><label>/<priority><right sub-treap>). The sub-treaps are printed recursively, and omitted if leafs.

Sample Input

7 a/7 b/6 c/5 d/4 e/3 f/2 g/1
7 a/1 b/2 c/3 d/4 e/5 f/6 g/7
7 a/3 b/6 c/4 d/7 e/2 f/5 g/1
0

Sample Output

(a/7(b/6(c/5(d/4(e/3(f/2(g/1)))))))
(((((((a/1)b/2)c/3)d/4)e/5)f/6)g/7)
(((a/3)b/6(c/4))d/7((e/2)f/5(g/1)))


题意:每一个节点有两个属性,“标号-优先级”,相对于标号是二叉搜索树,相对与优先级是大根堆 然后中根遍历

分析: 对标号进行从小到大排序,此时已是中根遍历了,然后在构造堆,插入的节点如果比他的父节点优先级小就放在右儿子位置。如果大的话就往上找直到找打比他大的那个节点记作x,此时x原来的右儿子更新为该节点的左儿子(中序遍历左儿子在前),该节点成为x的新右儿子。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int INF = 100000000;
struct node
{
    int lson,rson,fa;
    int v;
    char s[100];
};
node tree[50010];
int cmp(node x,node y)
{
    return strcmp(x.s, y.s) < 0;    //字符串比较用这个;换成return x.s < y.s 就WA了
}
void inset_node(int now)
{
    int j = now - 1;
    while(tree[j].v < tree[now].v)
        j = tree[j].fa;
    tree[now].lson = tree[j].rson;
    tree[j].rson = now;
    tree[now].fa = j;
}
void travel(int now)
{
    if(now == 0) return ;
    printf("(");
    travel(tree[now].lson);
    printf("%s/%d",tree[now].s,tree[now].v);
    travel(tree[now].rson);
    printf(")");
}
int main()
{
    int n;
    while(scanf("%d", &n) != EOF && n)
    {
        getchar();
        for(int i = 1; i <= n; i++)
        {
            scanf("%[a-z]/%d",tree[i].s,&tree[i].v);
            getchar();
            tree[i].lson = tree[i].rson = tree[i].fa = 0;
        }
        sort(tree + 1,tree + n + 1, cmp);
        tree[0].fa = tree[0].lson = tree[0].rson = 0;
        tree[0].lson = 0;
        tree[0].v = INF;
        for(int i = 1; i <= n; i++)
            inset_node(i);

        travel(tree[0].rson);
        printf("\n");
    }

    return 0;
}


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